Difference between revisions of "Solutions 6"

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(Nuclear reactions and nuclear reactors)
(Written and developed by Prof. Per Hoff (uio) )
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====== Written and developed by [http://www.mn.uio.no/kjemi/english/people/aca/phoff/index.html Prof. Per Hoff] (uio)   ======  
 
====== Written and developed by [http://www.mn.uio.no/kjemi/english/people/aca/phoff/index.html Prof. Per Hoff] (uio)   ======  
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Revision as of 10:01, 22 June 2012

Nuclear Reactions and Nuclear Reactors

Written and developed by Prof. Per Hoff (uio) 

Return to Problem Solving Sets


1: It is noteworthy to notice the Q-value for the neutron capture and the change in binding energy per nucleon for each of the isotope pairs, see table 6.2.

Table 6.2: Calculated Q-values and change in binding energy per nukleon
Pair of nuclide
Q-value for neutron capture (MeV)
Change in EB/A (MeV)
235U/236U
6.55
-0.004
238U/239U
4.81
-0.012
239Pu/240Pu
6.53
-0.003


The nuclide pair 238U/239U have a significantly lower Q-value and a significantly bigger fall in EB/A than the other pairs. This can be explained by the pair-pair configuration in the 238U nucleus, which makes it less favorable to bind another neutron. On the other hand, for pair-odd nuclides it is much more favorable to bind another neutron to achieve a pair-pair configuration. This is shown from the cross sections for interaction with thermal neutrons (σ and σf).


2: 

  1. 239Pu+ n [math]\rightarrow[/math] 99Y + 2n + 139Cs
  2. Q-value: 191.42MeV
  3. The energy which is released by disintegration after stability is reached:                        99Y: M(99Y)-M(99Ru)=17.4MeV                                                                                       139Cs: M(139Cs)-M(139La)=6.5MeV
  4. 2/3 of this energy will disappear with neutrinos. Some of the disintegrations have too long half-lives to have an effect on the reactor safety.


  1. 1.0g 239Pu = 2.5 • 1021 atomer. Number of fissions per seconds is σ • ϕ • Nt = 1.89 • 1014, which will give an effect of 3.6 • 1016MeV (5811W)
  2. The formation of 240Pu: σ • ϕ • Nt= 6.8 • 1013s-1. After 100 days of irradiation 4 • 10-6 g Pu will be made.

3:


  1. 232T'h+ η [math]\rightarrow[/math] 233Th [math]\rightarrow[/math] 233Pa [math]\rightarrow[/math] 233U
  2. 133I.
  3. One ton 232Th equals to 2.6*1027 atoms. The rate of formation for neutron capture (233Th): σ • ϕ • Nt = 7.37 • 1024cm2 • 1014n cm-2s-1 • 2.6 • 1027atomer= 1.91 • 1018atoms s-1
  4. It will take 37hours of irradiation to form enough 233Th to give 100g 233U, but disintegration of 233Pa to 233U must be waited.
  5. 100g 233U: D=λN = 3.56 • 1010Bq(35.6Gbq)